Answer
$$\frac{64}{3}$$
Work Step by Step
Given $$y=4-x^2,\ \ \ y=x^2-4 $$
First, we find the intersection points
\begin{align*}
4-x^2&=x^2-4\\
2x^2-8&=0
\end{align*}
Then $x=\pm 2$ and $ 4-x^2\geq x^2-4$ for $-2\leq x\leq 2$; hence, the area is given by
\begin{aligned}
\text{Area}&=\int_{-2}^{2}\left[\left(4-x^{2}\right)-\left(x^{2}-4\right)\right]d x\\
&=\int_{-2}^{2}\left(8-2 x^{2}\right) d x\\
&=2 \int_{0}^{2}\left(8-2 x^{2}\right) d x\\
&=2\left(8 x-\frac{2}{3} x^{3}\right)\bigg|_{0}^{2}\\
&=\frac{64}{3}
\end{aligned}
We use the Mathematica program to plot the region.
